Can anyone solve for "x"?

[farmerjohn1324]

ya = 0.2xa * (0.8x + c + w)

 

[Ishuda]

ya = 0.2xa * (0.8x + c + w)

Are ya and xa each variables or does it stand for y*a and x*a. If the latter (and a is not zero), you can divide through by a to eliminate it. Depending on that answer, you have either a linear equation in x, i.e. an
A x + B = 0,
or a quadratic equation, i.e.
A x² + B x + C = 0.
where the A, B, and C 'constants' will depend on the 0.2, y, c, and w and/or possibly xa and ya.

 

[farmerjohn1324]

ya, xa

Are ya and xa each variables or does it stand for y*a and x*a. If the latter (and a is not zero), you can divide through by a to eliminate it. Depending on that answer, you have either a linear equation in x, i.e. an
A x + B = 0,
or a quadratic equation, i.e.
A x² + B x + C = 0.
where the A, B, and C 'constants' will depend on the 0.2, y, c, and w and/or possibly xa and ya.

ya = y*a xa = x*a

a is not zero.

So what did you get?

I am trying.

Is it even possible to solve for x?

 

[Ishuda]

ya = y*a xa = x*a

a is not zero.

So what did you get?

I am trying.

Is it even possible to solve for x?

Since ya=y*a and xa=x*a and a is not zero we canwrite
ya = a * y = 0.2xa * (0.8x + c + w) = a * [0.2 x (0.8 x + c + w)]
or, dividing by a on both sides of the equation
y = 0.2 x (0.8 x + c + w) = 0.16 x² + 0.2 (c + w) x
or
0.16 x² + 0.2 (c + w) x - y = 0
which is a quadratic equation in x. If you don't know about quadratic equations and how to solve them then you either need to review them or ask your instructor to not give you problems you have not studied.

 

[farmerjohn1324]

quadratics

Since ya=y*a and xa=x*a and a is not zero we canwrite
ya = a * y = 0.2xa * (0.8x + c + w) = a * [0.2 x (0.8 x + c + w)]
or, dividing by a on both sides of the equation
y = 0.2 x (0.8 x + c + w) = 0.16 x² + 0.2 (c + w) x
or
0.16 x² + 0.2 (c + w) x - y = 0
which is a quadratic equation in x. If you don't know about quadratic equations and how to solve them then you either need to review them or ask your instructor to not give you problems you have not studied.

I know about quadratics.

ax² + bx + c = 0

Unfortunately, I'm not in school anymore, so I don't have an instructor.

I haven't "factored" since I was in 10th grade.

I'm going to try, but I am so out of practice, I don't want to mess it up.

Do you mind helping me with this?

We could at least compare answers. Trust me, this equation was simplified from a much more complicated one. I did as much as I could. I will try, though.

 

[farmerjohn1324]

quadratic

I am having trouble equating 0.16 x² + 0.2 (c + w) x - y = 0

... to the quadratic formula... ax^2 + bx + c = 0

I wouldn't even know where to begin trying to factor that.

 

[Deleted member 4993]

I am having trouble equating 0.16 x² + 0.2 (c + w) x - y = 0

... to the quadratic formula... ax^2 + bx + d = 0 (I used 'd' here to avoid confusion - because your original equation has a 'c' )

I wouldn't even know where to begin trying to factor that.

a = 0.16

b = 0.2*(c+w)

d = -y (I used 'd' here to avoid confusion - because your original equation has a 'c' )

Now use

\(\displaystyle \displaystyle{x_{1,2} \ = \ \frac{-b \pm \sqrt{b^2-4ad}}{2a} } \)

Now substitute a, b & d and calculate x.

 

[farmerjohn1324]

x

a = 0.16

b = 0.2*(c+w)

d = -y (I used 'd' here to avoid confusion - because your original equation has a 'c' )

Now use

\(\displaystyle \displaystyle{x_{1,2} \ = \ \frac{-b \pm \sqrt{b^2-4ad}}{2a} } \)

Now substitute a, b & d and calculate x.

Except for the fact that "b" contains two variables in itself.

Can I be honest with you? I'm slightly drunk, and I haven't done this since 10th grade.

What did you get?

Maybe I could try it tomorrow, but I would still have to check with you to make sure it's right.

And also...

From the original equation....

a = (F23*0.2)*((((G$7/12)*(1+(G$7/12))^G$6))/(((1+(G$7/12))^G$6)-1))

w = ((F23*0.8)+10000)*((((H$7/12)*(1+(H$7/12))^(H$6*12)))/(((1+(H$7/12))^(H$6*12))-1))

I greatly reduced those formulas. That might change things. I think I have to tediously reduce things to those complex formulas to solve this one.

Are you sure that it's a quadratic and that I don't have some circular logic somewhere, due to the fact that I reduced "a" and "w" to those simple variables?

 

[farmerjohn1324]

x....

NOTE: that's the quadratic equation, not the quadratic formula;
formula is x = [-b +- SQRT(b^2 - 4ac)] / (2a) ; remember?

Multiply by 100:
16x^2 + 20(c + w)x - 100y = 0
Divide by 4:
4x^2 + 5(c + w)x - 25y = 0

So we have a = 4, b = 5(c + w), c = -25y

Ya'll ok now

I'm not okay. What is "x"?

I need to plug this into my MS Excel file.

 

[farmerjohn1324]

booze...

What? "x" is part of YOUR initial equation:
"ya = 0.2xa * (0.8x + c + w)"

Plus here's YOUR question: "Can anyone solve for "x"?

Put the booze down, get some shuteye, and come back tomorrow.
We don't know what you mean when you say you've reduced
initial equations...SHOW us!!

I'm not that drunk. I still know what I'm trying to figure out.

https://www.dropbox.com/s/rg7gzlwysjk87y5/formula.xls?dl=0

There is the Excel file.

You can see that F9:I23 represents figuring out how much of a certain amount of loan I can take out.
"G" column represents a personal loan at 10.75% over 84 months.
"H" column represents a commercial real estate loan of 4.5% over 20 years.

"I" column is the sum of the monthly totals of these two loan payments.

I am trying to figure out that same formula... but do it backwards... so that I can figure out exactly how much of a loan I can take out based
on my "income minus expenses," which is represented by cells "C4:C6." C6 being the figure that is how much extra I have left over after my
expenses. ($2193.90).

The goal is to be able to put a figure in "F24" that figures out the formula in reverse and tells me how much the personal loan payment would be,
as well as the commercial real estate loan payment.

The plan is to finance to real estate purchases with 20% down payments being financed by 10.75% personal loans over 84 months. And I would also like to leave the variable that are in "G6:H7" open in case I need to change them.

a = (F23*0.2)*((((G$7/12)*(1+(G$7/12))^G$6))/(((1+(G$7/12))^G$6)-1))

w = ((F23*0.8)+10000)*((((H$7/12)*(1+(H$7/12))^(H$6*12)))/(((1+(H$7/12))^(H$6*12))-1))

..... as represented in G23 and H23.

See.... not that drunk. I was exaggerating. I think my original question was impossible because it involved circular logic due to the fact that I didn't try to factor in the value of "a" and "w" that is in here.

 

[farmerjohn1324]

excel...

Except that when I just tried to download the Excel file I just uploaded, it didn't have the formulas, it just had the values.

Does anyone know how I can share this file with you, other than through personal email?

I would gladly upload it to any service available.

 

[farmerjohn1324]

Sheet4

Okay, nevermind, I got it.

https://www.dropbox.com/s/cujjmsh29dm26uj/home numbers.xlsx?dl=0

Here is the new file.

On the tab called "Sheet4," which is the first tab of the file.

 

[Ishuda]

I'm not okay. What is "x"?

I need to plug this into my MS Excel file.

Lets go through a simple example. First, when we reduced your problem to a quadratic equation we obtained
0.16 x² + 0.2 (c + w) x - y = 0
Now, suppose you put some numbers in your spread sheet and got the following results:
c = 10,
w = -8.4,
and
y = 1.28.
Then the equation becomes
0.16 x² + 0.2 (10 - 8.4) x - 1.28 = 0.16 x² + 0.2 * 1.6 x - 1.28 = 0.16 x² + 0.32 x - 1.28 = 0
If we divide through by 0.16 we get
x² + 2 x - 8 = 0.
Now use the quadratic formula as given above by Denis
x = [-B +- SQRT(B^2 - 4AC)] / (2A)
where A=1, B=2, and C=-8. Thus
x = \(\displaystyle \dfrac{-2\, \pm\sqrt{2^2-4(1)(-8)}}{2(1)}\, =\, \dfrac{-2\, \pm\sqrt{4+32}}{2}\, \)
=\(\displaystyle \dfrac{-2\, \pm\sqrt{36}}{2}\, =\, \dfrac{-2\, \pm\, 6}{2}\, =\, -1\, \pm\, 3\)
or x is either -4 or 2.

Now you can either wait until you get some actual numbers and just plug them in or you can write the equation symbolically such as given by Subhotosh Khan or Denis.

 

[farmerjohn1324]

numbers....

Lets go through a simple example. First, when we reduced your problem to a quadratic equation we obtained
0.16 x² + 0.2 (c + w) x - y = 0
Now, suppose you put some numbers in your spread sheet and got the following results:
c = 10,
w = -8.4,
and
y = 1.28.
Then the equation becomes
0.16 x² + 0.2 (10 - 8.4) x - 1.28 = 0.16 x² + 0.2 * 1.6 x - 1.28 = 0.16 x² + 0.32 x - 1.28 = 0
If we divide through by 0.16 we get
x² + 2 x - 8 = 0.
Now use the quadratic formula as given above by Denis
x = [-B +- SQRT(B^2 - 4AC)] / (2A)
where A=1, B=2, and C=-8. Thus
x = \(\displaystyle \dfrac{-2\, \pm\sqrt{2^2-4(1)(-8)}}{2(1)}\, =\, \dfrac{-2\, \pm\sqrt{4+32}}{2}\, \)
=\(\displaystyle \dfrac{-2\, \pm\sqrt{36}}{2}\, =\, \dfrac{-2\, \pm\, 6}{2}\, =\, -1\, \pm\, 3\)
or x is either -4 or 2.

Now you can either wait until you get some actual numbers and just plug them in or you can write the equation symbolically such as given by Subhotosh Khan or Denis.

The problem is that the variables are not constants, and in the spread sheet, they have to be able to be changed at a moment's notice, so they kind of have to remain as variables in a way.

 

[farmerjohn1324]

formula...

Had a look...I see what you're trying for...
I'll give you a solution MY WAY; not familiar with Excel,
plus will never be: hate it more every time I see it :shock:

GIVENS:
i = interest pers. loan (.1075 / 12)
j = interest com. loan (.045 / 12)
m = months pers. loan (84)
n = months com. loan (240)
d = disposable (2194) ......... I added a dime, ok?

Step 1 (a = payment factor pers. loan):
a = [1 - 1/(1 + i)^m] / i

Step 2 (b = payment factor com. loan):
b = [1 - 1/(1 + j)^n] / j

Now we're ready for what you can borrow!
pers. loan payment = p = bd / (4a + b) : that'll be 881

com. loan payment = c = d - p : that'll be 1,313

pers. loan amount = u = a*p : that'll be 51,871

com. loan amount = v = 4*u : that'll be 207,484

If you're able to set that up properly,
then you'll be able to calculate at any rates,
at any terms in months, at any disposable income.

Thanks!!!

 

[Ishuda]

The problem is that the variables are not constants, and in the spread sheet, they have to be able to be changed at a moment's notice, so they kind of have to remain as variables in a way.

You have cells where c, w, and y are computed with a formula or just by changing the values. You have other cells which will contain the A, B, and C. The cell for A is the constant 0.16, the cell for B is the formula 0.5*(c+w) where c and w are actually the cell addresses for c and w, i.e. $c$4 and $d$4 for example. The cell for C is the formula -y where, as before, the y is actually the cell address for y. Now use one cell for the + in the quadratic formula and another for the - to get the two answers. Note: it may be possible the B²-4AC becomes negative, in that case you would get an error or you could program around it and get some default value, say some ridiculous number so you know there is no solution.

 

[farmerjohn1324]

circular reference

Thanks!!!

Your formula contains a circular reference when calculating "u" and "v."

I think the problem lies in that calculating "c" should be more complicated than just subtracting "p" from "d."

Do you know how to fix this?

And... punching in the numbers manually....

The personal loan payment should come to $855.899 and the commercial loan should be $1338.00, and the total loan I can afford is $251864.52.